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\notesname{Notes:}
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% Ok, I get sky, and I get pink, but couch?

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\begin{document}
\title{California Blues}
\author{Brigitte Servatius}
\maketitle





\begin{slide}
\section{Geometric rigidity on the fixed torus}

We are interested in three objects:
\begin{enumerate}
	\item An infinite periodic graph, realized in $\mathbb R^2$ (infinite periodic framework).

	\item Its underlying {\it gain graph:} $\langle G, g \rangle$, 
where $G$ is the quotient graph under translational symmetry, and $g$ is a labeling of the 
edges by elements of the integer lattice $\mathbb Z^2$

	\item The {\it geometric gain framework}: The quotient graph $G$ realized in $\mathbb R^2$ 
with same vertex geometry as in the infinite periodic graph, but now with straight edges, and no gains.
\end{enumerate}

Notation: Denote the infinite periodic framework as $(\langle G, g \rangle, p)$, and 
denote the geometric gain framework by $(G,p)$.

\begin{theorem}
Fix one vertex of the infinite periodic framework, and fix the corresponding vertex of the geometric gain graph. 
If there is a nontrivial (periodic) infinitesimal motion of the infinite periodic framework 
$(\langle G, g \rangle, p)$, then we can ``shrink" this down to an infinitesimal motion of the geometric gain 
framework $(G,p)$. The infinitesimal motion of the geometric gain framework is either a rotation about the fixed 
vertex, or is non-trivial.
\end{theorem}

In other words, every motion of the infinite periodic framework corresponds to an infinitesimal motion of the 
geometric gain framework.
\end{slide}
\newpage

\begin{slide}
Ideas for the proof:
\begin{itemize}
	\item We know how to do certain cases:
    	\begin{enumerate}
        	\item only one edge has a gain
	\item The zig-zag graph: two edges, two vertices
         \item (?) only one gain, say $(a, b)$, on multiple edges
         \item (?) graphs that are not two-connected, where the gains appear in separate components
         \item (?) two edges with two gains, as in the $K_4$ example.
        \end{enumerate}
    \item We can always assume that there is a spanning tree of edges with zero gains. Using that, the cycle space has a basis with one basis cycle corresponding to each non-tree edge.
    \item Can we view all gains as linear combinations of two independent gains? Here ``independent'' means both linearly independent as vectors in $\mathbb Z^2$ and also independent as cycles.
        \item we can consider ``families'' of edges, where the edges are arranged into families according to their gains (think of these as colours)
        \item what insight do we get by considering the realization where all vertices have the same position?

\end{itemize}

\end{slide}
\newpage

\begin{slide}
\begin{conjecture}
If there is a nontrivial infinitesimal motion of the geometric gain graph, 
it may lift (under some conditions: what are they?) to a nontrivial (periodic) infinitesimal motion of the 
infinite periodic graph on the fixed torus.
\end{conjecture}

Notes about this direction:
\begin{itemize}
	\item each cycle gives us some nontrivial condition on the lifted motion
    \item each edge with a non-trivial gain will correspond to an implicit triangle in the lifted realization - 
    we know how to lift the motion

\end{itemize}
\end{slide}
\newpage

\begin{slide}

\section{Notation}

\begin{itemize}
	\item multigraph $G = (V, E)$, with $|V| = n$, and $|E| = m$
	\item gain graph $\langle G, g \rangle$, where $g$ is the gain assignment, defined below
	\item vertices $v_1, \dots, v_n$
	\item configuration $p = (p_1, \dots, p_n)$, where $p_i \in \mathbb R^2$, $p_i = (p_{i1}, p_{i2})$
	\item edges $e_1, \dots, e_m$
	\item gains $g_1, \dots, g_m$, where $g_k \in \mathbb Z \times \mathbb Z$
	\item an edge of the gain graph is denoted $e_k = \{v_i, v_j; g_k\}$, and this says that the edge is originating in vertex $v_i$, terminating in vertex $v_j$, and is labeled with the gain $g_k$. This edge is equivalent to $\{v_j, v_i; -g_k\}$.
	\item infinitesimal velocity assignment $u = (u_1, \dots u_n)$, where $u_i \in \mathbb R^2$.
	\item periodic orbit framework on the (fixed) torus is $(\langle G, g \rangle, p)$.
\end{itemize}

\end{slide}
\newpage

\begin{slide}
\section{Definitions}
\begin{itemize}
	\item define {\it shrink}
\end{itemize}

The  rigidity matrix $R_{G}(p)$ for a finite graph is:
\[
\begin{blockarray}{cccccc}
& \cdots & v_i & \cdots & v_j & \cdots \\
\begin{block}{c[ccccc]}
 \vdots & & & & & \\
e_{ij} = (v_i, v_j) & 0 \cdots 0 & p_i - p_j & 0 \cdots 0 & p_j - p_i & 0 \cdots 0 \\
   \vdots & & & & & \\
\end{block}
\end{blockarray}
 .\]
\end{slide}
\newpage

\begin{slide}
The (fixed torus) periodic rigidity matrix $R_{\langle G, g \rangle}(p)$ is:
\[
\begin{blockarray}{cccccc}
& \cdots & v_i & \cdots & v_j & \cdots \\
\begin{block}{c[ccccc]}
 \vdots & & & & & \\
 e_k =  \{v_i, v_j; g_k\} & 0 \cdots 0 & p_i - (p_j+ g_e) & 0 \cdots 0 & (p_j + g_e) - p_i & 0 \cdots 0 \\
   \vdots & & & & & \\
\end{block}
\end{blockarray}
 .\]

For a gain graph $\langle G, g \rangle$, the {\it gain matrix} $R_g$ is:
  \[
\begin{blockarray}{cccccc}
& \cdots & v_i & \cdots & v_j & \cdots \\
\begin{block}{c[ccccc]}
 \vdots & & & & & \\
 e_k =  \{v_i, v_j; g_k\} & 0 \cdots 0 & - g_e & 0 \cdots 0 & g_e & 0 \cdots 0 \\
   \vdots & & & & & \\
\end{block}
\end{blockarray}
 .\]
Note that the gain matrix is independent of $p$.
\end{slide}

\begin{slide}
 Suppose that $u$ is an infinitesimal motion of $R_{\langle G, g \rangle}(p)$ on the fixed torus. Then $R_{\langle G, g \rangle}(p)\cdot u = 0$. We may rewrite
 \[R_{\langle G, g \rangle}(p)\cdot u = R_G(p)\cdot u + R_g\cdot u = 0,\]
 where $R_G(p)$ is the rigidity matrix for the geometric gain graph.

 In the situation where the infinitesimal motion assignment $u$ on the periodic orbit framework is also an infinitesimal motion assignment on the geometric gain graph, it must be the case that $R_g\cdot u = 0$.
\end{slide}

\begin{slide}
\section{Examples}

\subsection{Zig-zag graph}
We begin with a small example. Let $G$ be the graph with two vertices, $v_1, v_2$ and two edges, $e_1 = \{v_1, v_2; (0,0)\}$, and $e_2 = \{v_1, v_2; g\}$, where $g \in \mathbb Z^2$. Let $p_1$ and $p_2$ represent the coordinates of the vertices $v_1$ and $v_2$ on the fixed unit torus $\mathcal T_0^2$. Then the rigidity matrix for this framework on $\mathcal T_0^2$ is the following:
\[
\begin{blockarray}{ccc}
 & v_1 & v_2  \\
\begin{block}{c[cc]}
  e_1 & p_1 - p_2 & p_2 - p_1 \\
  e_2 & p_1 - (p_2+g) & (p_2 +g) - p_1  \\
\end{block}
\end{blockarray}
 .\]

This row reduces to
\[
\begin{blockarray}{ccc}
 & v_1 & v_2  \\
\begin{block}{c[cc]}
  e_1 & -(p_2 - p_1) & p_2 - p_1 \\
  e_2 & -g & g  \\
\end{block}
\end{blockarray}
 ,\]
 and we see immediately that the framework has a non-trivial infinitesimal 
 motion if and only if $p_2 - p_1$ is in the same direction as the gain $g$. 
 In other words, both edges lie on an single line.

\end{slide}

\begin{slide}
$$
{\begin{tikzpicture}[->,>=stealth,shorten >=1pt,auto,node distance=2.8cm,thick, font=\footnotesize]

\tikzstyle{vertex1}=[circle, draw, fill=couch, inner sep=1pt, minimum width=3.5pt];
\tikzstyle{vertex2}=[circle, draw, fill=lips, inner sep=1pt, minimum width=3.5pt];
\tikzstyle{vertex3}=[circle, draw, fill=melon, inner sep=1pt, minimum width=3.5pt];
\tikzstyle{vertex4}=[circle, draw, fill=bluey, inner sep=1pt, minimum width=3.51pt];
\tikzstyle{gain} = [fill=white, inner sep = 3pt,  font=\footnotesize, anchor=center];

\node[vertex1] (1) at (0, 0){$v_1$};
\node[vertex1] (2) at (1, 2) {$v_2$};

\path (1) edge [bend right] node[gain] {$g$} (2);
\path (1) edge [bend left] node[gain] {$(0,0)$} (2);

\end{tikzpicture}} \hspace{1cm}%
%
% Two vertices on the torus
{\begin{tikzpicture}
\tikzstyle{vertex1}=[circle, draw, fill=couch, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex2}=[circle, draw, fill=lips, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex3}=[circle, draw, fill=melon, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex4}=[circle, draw, fill=bluey, inner sep=1pt, minimum width=4pt];

\node[vertex1] (1) at (1,-1) {};
\node[vertex1] (2) at (0,0) {};

\draw[thick] (1) -- (2)  ;
\draw[thick] (1) -- (1.5, -0.5);
\draw[thick] (-0.5, -0.5) -- (2);

\pgfsetarrowsend{latex}
	\draw[sky, very thick] (-.5, -1.5) -- (1.5, -1.5);
	\draw[sky, very thick] (-.5, .5) -- (1.5, .5);
	\draw[sky, very thick] (-.5, -1.5) -- (-.5, .5);
	\draw[sky, very thick] (1.5, -1.5) -- (1.5, .5);
\pgfsetarrowsend{}
\end{tikzpicture}}\hspace{1cm}%
{\begin{tikzpicture}
\tikzstyle{vertex1}=[circle, draw, fill=couch, inner sep=1pt, minimum width=3pt];
\tikzstyle{vertex2}=[circle, draw, fill=lips, inner sep=1pt, minimum width=3pt];

\foreach \x in {-2, -1, 0,1,2}
\foreach \y in { -1, 0,1}
{
\node[vertex1] (1\x\y) at (\x+.5, \y-.5){};
}

\foreach \x in {-2, -1, 0,1,2}
\foreach \y in {-1, 0,1}
{
\node[vertex1] (2\x\y) at (\x,\y) {};
}
%Edges
\draw \foreach \x in { -2,-1, 0,1, 2}
\foreach \y in {-1, 0,1} {(1\x\y) -- (2\x\y)};

\draw \foreach \x in { -2,-1, 0,1,2}
\foreach \y in {-1, 0,1} {(1\x\y) -- (\x+1, \y) };

Redrawing vertex 1 for prettiness
\foreach \x in {-2, -1, 0,1,2,3}
\foreach \y in { -1, 0,1}
{
\node[vertex1] (2\x\y) at (\x, \y) {};
}

%\pgfsetarrowsend{latex}
	\draw[sky, very thick] (-.25, -.75) -- (.75, -.75);
	\draw[sky, very thick] (-.25, .25) -- (.75, .25);
	\draw[sky, very thick] (-.25, -.75) -- (-.25, .25);
	\draw[sky, very thick] (.75, -.75) -- (.75, .25);
	%\pgfsetarrowsend{}

\end{tikzpicture}}\hspace{1cm}%
{\begin{tikzpicture}
\tikzstyle{vertex1}=[circle, draw, fill=couch, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex2}=[circle, draw, fill=lips, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex3}=[circle, draw, fill=melon, inner sep=1pt, minimum width=4pt];
\tikzstyle{vertex4}=[circle, draw, fill=bluey, inner sep=1pt, minimum width=4pt];

\node[vertex1] (1) at (1,-.5) {};
\node[vertex1] (2) at (0,-.5) {};

\draw[thick] (1) -- (2)  ;
\draw[thick] (1) -- (1.5, -0.5);
\draw[thick] (-0.5, -0.5) -- (2);

\pgfsetarrowsend{stealth}
\draw[pink, very thick] (1) -- (1,0);
%\draw[pink, very thick] (2) -- (0, -1);
\pgfsetarrowsend{}


\pgfsetarrowsend{latex}
	\draw[sky, very thick] (-.5, -1.5) -- (1.5, -1.5);
	\draw[sky, very thick] (-.5, .5) -- (1.5, .5);
	\draw[sky, very thick] (-.5, -1.5) -- (-.5, .5);
	\draw[sky, very thick] (1.5, -1.5) -- (1.5, .5);
\pgfsetarrowsend{}
\end{tikzpicture}}
$$
 We can also make this argument geometrically. We need to develop notation for the graph with the graph with the gain edges inserted. Note that these ``gain edges" do more than they advertise in the sense that we require a {\it periodic} motion on the vertices.
\end{slide}

\begin{slide}

\subsection{Generically flexible}
A gain graph $\langle G, m \rangle$ is generically flexible on the fixed torus if one of
\begin{enumerate}
	\item $|E(G)| \leq 2|V(G)| - 2$, or
	\item $|E(G) = 2|V(G)| - 2$ and there is some subgraph $G' \subset G$ with $|E(G')| = 2|V(G')| - 2$ with no constructive cycles (no cycle with non-trivial net gain).
\end{enumerate}

Since these are generically flexible, it should be obvious that there is a non-trivial 
infinitesimal motion at every position of the vertices. However, how do we show that we 
can transform these motions continuously to a non-trivial motion of the geometric gain graph?
\end{slide}

\begin{slide}


\subsection{Generically rigid, geometrically flexible: $K_4$}
By cite{Ross} this is generically rigid on the fixed torus. In this geometric position, it is flexible. We show that the infinitesimal motion on the periodic framework can be shrunken to an infinitesimal motion of the geometric gain graph.

Let the vertices be placed as follows:
\[p_0 = (0.5, 0), \  p_1 = (0,0), \ p_2 = (0, 0.5), \ p_3 = (0.5, 0.5).\]
Let $g_1 = (1,0)$, $g_2 = (-1,0)$.
The initial motion assignment $u$ is:
\[u = (\begin{array}{cccccccc} 0 & 0 & 0 & -a & a & -a & a & 0 \end{array}).\]
Or in other words:
\[u_0 = (0,0), \ u_1 = (0, -a), \ u_2 = (a, -a), \ u_3 = (a, 0).\]

We shrink this motion as follows: for $t \in [-1,1]$, let $g_1$ and $g(2)$ be written in homogeneous coordinates as $g_1(t) = [\frac{1-t}{2}: 0 :1]$ and $g_2(t) = [0: \frac{1-t}{2} : t]$. Note that as $t$ passes through zero, the $y$ coordinates in $g_2(t)$ passes through infinity. If we demand that $u_0(1) = u_0(-1)$, and $u_3(1) = u_3(-1)$ (i.e. that the initial velocities at the vertices $v_0$ and $v_3$ remain fixed), then as we shrink the gains, the infinitesimal motions at the vertices $v_1$ and $v_2$ are as follows:
\[u(t) = (\begin{array}{cccccccc} 0 & 0 & 0 & at & a & at & a & 0 \end{array}).\]
In particular, when $t=1$ and $g_1(1) = g_2(1) = (0,0)$, then
\[u(1) = (\begin{array}{cccccccc} 0 & 0 & 0 & a & a & a & a & 0 \end{array}),\]
which is an infinitesimal rotation of the geometric gain graph about the fixed vertex $p_0$.

\end{slide}

\begin{slide}
\section{More ideas and extensions: }
\begin{itemize}
	\item We can do this for the flexible torus too (question: how do we distinguish fixed torus motions from flexible torus motions?)
    \item The same should be true for graphs with different automorphism groups. For example, finite graphs with rotational symmetry. In this setting there are some additional considerations -- we may have vertices or edges that are fixed by the symmetry (e.g. vertices or edges on the axis of rotation or mirror of reflectional symmetry).
\end{itemize}

\end{slide}

\begin{slide}

\section{Example}
Consider the following graph, a quadrilateral with two opposite parallel edges, or dipoles.
$$\includegraphics{fourbar1aquotient} \qquad \includegraphics{fourbar1a}$$
In a generic embedding $p:V \rightarrow \mathbb{R}^2$ into the plane with one vertex pinned at the origin, it has two degrees of freedom, one of which is an
infinitesimal rotation, often considered trivial.  Both of these infinitesimal motions lift to the covering graph if
the group labels on the edges are all the identity, since each connected component will be isometric to $p$.

If one edge in one dipole has a non-trivial group label, then, usually $p$ lifts to a covering framework
which has a rigid infinite path, with subpaths of length two joining the pinned vertices, indicated in black.
$$\includegraphics{fourbar1bcover}$$

\end{slide}

\begin{slide}
If, however, the non-trivial group label for the edge $(i,j)$ is parallel to $p(i) - p(j)$,
then the entire space of infinitesimal motions lifts to the covering framework.
$$\includegraphics{fourbar1dcover}$$


It may happen each of the dipoles has one non-trivial group label, and that these elements are linearly dependent in
$\mathbb{Z}\times \mathbb{Z}$.
In this case the group action on the cover is only $\mathbb{Z}$ and it can happen that no infinitesimal motions lift,
as in the following case in which the group labels are both $\alpha \in \mathbb{Z}\times\mathbb{Z}$.
$$\includegraphics{fourbar1c4degcover}$$
It may also happen that neither of the infinitesimal motions lift if not only are the two group labels dependent, in
fact identical, in $\mathbb{Z}\times\mathbb{Z}$, but the vectors $p_i - p_j - \alpha$ are are parallel:
$$\includegraphics{fourbar1cdegcover}$$
Here is another case in which $3\alpha = 2\beta$, so they are dependent, and the lifted edges
$p_1 - p_2 + \alpha$ and  $p_3 - p_4 + \beta$ are parallel, even though
$p_1 - p_2$ and $p_3 - p_4$ are not parallel.
$$\includegraphics{fourbar1c2degcover}$$

\end{slide}

\begin{slide}
Note that even if you start with two group labels $\alpha$ and $\beta$ which are linearly independent in $\mathbb{Z}\times\mathbb{Z}$,
one may still engineer the covering framework so that the lifted edge
$p_1 - p_2 + 2\alpha$ is parallel to $p_3 - p_4 + 3\beta$:
$$\includegraphics{fourbar1c1degcover}$$
In this case the covering framework still has full symmetry group $\mathbb{Z}\times\mathbb{Z}$, and,
at least in this case, neither of the infinitesimal motions lift to the cover.

It is also possible to have dependent $\alpha$ and $\beta$ such that the lifted edges allow all the infinitesimal
motions to lift.  Here $\alpha = \beta$ and  $p_1 - p_2$,  $p_3 - p_4$,   $p_1 - p_2 + \alpha$ and $p_3 - p_4 + \beta$
are all parallel.
And both motions lift.
$$\includegraphics{fourbar1c1cover}$$
and
$$\includegraphics{fourbar1c2cover}$$


\end{slide}

\begin{slide}
Finally, we can take $\alpha$ and $\beta$ to be the generators of $\mathbb{Z}\times\mathbb{Z}$, so clearly independent,
but by choosing the generators carefully, we can make sure that all the motions lift to the cover.
$$\includegraphics{fourbar1acover}$$
and we note that none of the lifted infinitesimal motions are trivial on the fixed torus.

\end{slide}

\begin{slide}
\section{Extreme Examples}
    Embedding the quotient graph with all the $n$ vertices at the same location.  The space of infinitesimal motions has, in the case, $2(n-1)$
    degrees of freedom (with one vertex fixed), because at each non-fixed vertex a velocity vector can be freely chosen.

\end{slide}

\begin{slide}
\subsection*{The Dipole with Dependent Gain Space}
    The dipole in this collapsed embedding has a two dimensional space of infinitesimal motions, with rotations counting.
    If one edge is assigned a non-zero gain vector, the resulting framework is a disjoint union of straight lines and has a
    one dimensional space of infinitesimal motions.  If the two edges are assigned gain vectors which are linearly independent, the resulting
    framework is a grid and is rigid.  See Figures~\ref{figg221c}
$$    \includegraphics[scale=.6]{g221}
$$ 
{The periodic framework is a superposition of a) and b).  The velocity vector is perpendicular to the gain vector.\label{figg221}}

$$
    \includegraphics{g221b} \qquad \includegraphics{g221c}
$$
The abstract quotient is a dipole, a).  The geometric quotient has a velocity vector of arbitrary direction.\label{figg221c}

\end{slide}

\begin{slide}

\subsection*{The Dipole with Independent Gains}
    Here the infinitesimal velocities on the pinned periodic framework must be zero, however the geometric quotient framework
    still has two degrees of freedom. See Figures~\ref{figg223,figg223c}
$$    \includegraphics[scale=.6]{g223}$$
{The periodic framework is a superposition of a) and b).  The velocity vectors are all zero.\label{figg223}}

$$    \includegraphics{g223b} \qquad \includegraphics{g223c}$$
{The abstract quotient is a dipole, a).  The geometric quotient has a velocity vector of arbitrary direction.\label{figg223c}}

\end{slide}

\begin{slide}
\subsection*{Quotient cycles non-zero in gain space.}
If we take the quotient graph to consist of a triangle such that the sum of the grains around the oriented
$$    \includegraphics[scale=.6]{g229c}$$
{Quotient graph is a $3$-cycle. \label{figg229c}}

cycle is non-zero, then the periodic framework will have connected components consisting of infinite paths.
If $a$ is pinned, then the the velocities at $b$ and $c$ have to be perpendicular to the gain vectors, and must project equally on the
third gain vector, and so have equal length, and the infinitesimal degree of freedom is $1$.  See Figures~\ref{figg229c,figg229,figg229b}

$$    \includegraphics[scale=.6]{g229}$$
{The periodic framework is a superposition of a-d).\label{figg229} }

$$
    \includegraphics[scale=.6]{g229b}$$
{The periodic framework is a superposition of a-d).\label{figg229b}}

The geometric quotient has infinitesimal degree of freedom $2$.

\end{slide}

\begin{slide}
\subsection*{Quotient cycles zero in gain space.}
If we take the quotient graph to consist of a triangle such that the sum of the grains around the oriented
$$
    \includegraphics[scale=.6]{g230c}
$$
{Quotient graph is a $3$-cycle. \label{figg230c}}
cycle is zero, then the connected components of the lift will be triangles.  If we pin $a$ the conditions on
$b$ and $c$ are the same, and we again get the speeds must be the same, but now the motion center of the infinitesimal
rotation of the lift of the edge $(b,c)$ must be $a$.  See Figures~\ref{figg230c,figg230,figg230b}

$$    \includegraphics[scale=.6]{g230}$$
{The periodic framework is a superposition of a-d). \label{figg230}}

$$
    \includegraphics[scale=.6]{g230b}$$
{The periodic framework is a superposition of a-d).\label{figg230b}}

\end{slide}

\begin{slide}
\subsection*{Cover is connected.  Length 0 edge.}
The lifts of  vertices $a$ and $b$ have velocity $0$ as in the previous example.
$$
    \includegraphics[scale=.6]{g232} $$
{The periodic framework is a superposition of a-d). \label{figg232}}

$$
    \includegraphics[scale=.6]{g232b}$$
{The periodic framework is a superposition of a-d).\label{figg232b}}

Velocity at $c$ is perpendicular to gain vector $(1,1)$.  No other restriction.


\end{slide}

\begin{slide}
\subsection*{Cover is connected. Rigid.}
The dipole on $a$ and $b$ lifts to a rigid zig-zag path, as before.
$$
    \includegraphics[scale=.6]{g233}$$
{The periodic framework is a superposition of a-d). \label{figg233}}

The lift of the vertex $c$ serves to join the zig-zag paths together, as in the
previous example, so that the entire lift is connected, however here the lifts of
the vertex $c$ are also forced to have velocity zero.
$$
    \includegraphics[scale=.6]{g233b}$$
{The periodic framework is a superposition of a-d).\label{figg233b}}

\end{slide}

\begin{slide}
\section{Special Vertices}
    In the comparison and deformation of frameworks, it is often helpful to consider relaxing the specifications for a framework to include
    some limiting cases.  Specifically, each framework is an embedding $p$ of the vertices $V$ of a graph into $\mathbb{R}^n$.

    \subsection{Coincident vertices}  If we relax the condition that the function $p$ is an embedding, then it is possible for
    $p_v = p_{v'}$ for some pair of vertices.  For some graphs, the possibility would still be forbidden in the case of generic frameworks,
    certainly if $(v,v') \in E$, in which case $p_{v'} - p_{v} = \mathbb{0}$, and so the edge provides no constraint on the framework.

    If we are deforming a framework, when two vertices pass over one another at a finite velocity, there is a constraint at all times
    except the moment of coincidence.  In analyzing that moment, it is necessary to consider the constraint on the velocities from the
    deformation itself, which will be the limit
        $$   \lim_{t \rightarrow 0} \frac{(\dot{p}_v(t) - \dot{p}_v'(t))\cdot ({p}_v(t) - {p}_v'(t))}{|{p}_v(t) - {p}_v'(t)|} = 0$$
    which can be consider a constraint coming from an ``infinitesimal edge'' from $v$ from $v''$.


\end{slide}

\begin{slide}




A {\it gain graph} is a graph $G$ whose edges are labeled invertibly by the elements of a group $\Gamma$ \cite{Zaslavsky}. We denote a gain graph by $\langle G, g \rangle$, where $g: E(G)^+ \rightarrow \Gamma$ is the labelling of the forward-directed edges.
If the directed edge $e$ has label $g_e$, then the other direction of the edge has label $g_e^{-1}$ (see Figure \ref{fig:gain}).
When $\Gamma = \mathbb Z^2$, the gain graph $\langle G, g \rangle$, together with a {\it realization} $p$ of the vertices $V(G)$ in the plane, provides a description of an infinite periodic graph realized in $\mathbb R^2$ (see Figure 2).
We view $p$ as a realization of the vertices on the {\it fixed torus}: the topological torus formed by identifying the opposite sides of the square $[0,1] \times [0,1]$.
\end{slide}

\begin{slide}

The present work is concerned with three objects:
\begin{enumerate}
	\item An infinite periodic graph, realized in $\mathbb R^2$ ({\it infinite periodic framework}). Denote by $(\langle G, g \rangle, p)$.

	\item Its underlying {\it gain graph:} $\langle G, g \rangle$, where $G$ is the quotient graph under
translational symmetry, and $g$ is a labelling of the edges by elements of $\mathbb Z^2$.

	\item The {\it geometric gain framework}: The quotient graph $G$ realized in $\mathbb R^2$ with same
vertex geometry as in the infinite periodic graph, but now with straight edges, and no gains. Denote by $(G,p)$.
\end{enumerate}

%Notation: Denote the infinite periodic framework as $(\langle G, g \rangle, p)$, and denote the geometric gain framework by $(G,p)$.
\end{slide}

\begin{slide}
An {\it infinitesimal motion} of an infinite periodic framework is an assignment $u: V(G) \rightarrow \mathbb R^2$ of infinitesimal velocities to the vertices of the gain graph $\langle G, g \rangle$ on the fixed torus such that the edges lengths of the framework are (infinitesimally) preserved. If an infinitesimal motion preserves the distances between {\it all} pairs of vertices, it is called {\it trivial}. Such an infinitesimal motion is forced to be {\it periodic}, since it is defined on orbits of vertices and edges in the quotient graph.

\begin{theorem}
Fix one vertex of the infinite periodic framework, and fix the corresponding vertex of
the geometric gain graph. If there is a nontrivial infinitesimal motion of the
infinite periodic framework $(\langle G, g \rangle, p)$, then we can ``shrink" the  gains
to obtain an infinitesimal motion of the geometric gain framework $(G,p)$. The infinitesimal motion
of the geometric gain framework is either a rotation about the fixed vertex, or is non-trivial.
\end{theorem}

In other words, every infinitesimal motion of the infinite periodic framework corresponds to an infinitesimal motion
of the geometric gain framework. The converse is not true. However, it is an interesting question to determine which motions of the geometric gain graph
lift to motions of the infinite periodic framework.
\end{slide}

\begin{slide}

The following example of a generically rigid graph which is geometrically flexible provides an
illustration of our methods. By \cite{Ross} $K_4$ is generically rigid on the fixed torus. In the geometric position with coordinates given below,
it is flexible. We show that the infinitesimal motion on the periodic framework can be shrunken to an
infinitesimal motion of the geometric gain graph.

$$\includegraphics[scale=.65]{abgrid00a}  \qquad
\includegraphics[scale=.65]{abgrid00ax}$$
{Gain graph $\langle G, g \rangle$. Unlabelled edges have gain $(0,0)$. \label{fig:gain}}

$$\includegraphics[scale=.45]{abgrid02a} \quad \includegraphics[scale=.45]{abgrid01a}$$

{The periodic framework corresponding to the gain graph of Figure \ref{fig:gain} (left). The ``shrink" of this infinitesimal motion (red arrows) in progress (right). }
\end{slide}

\begin{slide}
Let the vertices be placed as follows:
\[
p_0 = (0.5, 0),
\  p_1 = (0,0),
\ p_2 = (0, 0.5),
\ p_3 = (0.5, 0.5).
\]
Let $g_1 = (1,0)$, $g_2 = (-1,0)$.
The initial motion assignment $u$ is:
\[u = (\begin{array}{cccccccc} 0 & 0 & 0 & -a & a & -a & a & 0 \end{array}).\]
Or in other words $u_0 = (0,0), \ u_1 = (0, -a), \ u_2 = (a, -a), \ u_3 = (a, 0)$.

We shrink this motion as follows: for $t \in [-1,1]$, let $g_1$ and $g_2$ be written in homogeneous
coordinates as $g_1(t) = [\frac{1-t}{2}: 0 :1]$ and $g_2(t) = [0: \frac{1-t}{2} : t]$. Note that as
$t$ passes through zero, the $y$ coordinate of $g_2(t)$ passes through infinity. If we demand that
$u_0(1) = u_0(-1)$, and $u_3(1) = u_3(-1)$ (i.e. that the initial velocities at the vertices
$v_0$ and $v_3$ remain fixed), then as we shrink the gains, the infinitesimal motions at the
vertices $v_1$ and $v_2$ are as follows:
\[u(t) = (\begin{array}{cccccccc} 0 & 0 & 0 & at & a & at & a & 0 \end{array}).\]
In particular, when $t=1$ and $g_1(1) = g_2(1) = (0,0)$, then
\[u(1) = (\begin{array}{cccccccc} 0 & 0 & 0 & a & a & a & a & 0 \end{array}),\]
which is an infinitesimal rotation of $(\langle G, g \rangle, p)$ about the fixed vertex $p_0$.

\end{slide}


\begin{thebibliography}{9}

\bibitem{Ross}
  Elissa Ross,
  \emph{Geometric and combinatorial rigidity of periodic frameworks as graphs
 on the torus}.
Ph.D. Thesis York University (Canada).
ProQuest LLC, Ann Arbor, MI,  2011. 336 pp. ISBN: 978-0494-75689-8

\bibitem{Zaslavsky}
T.~Zaslavsky.
\newblock Biased graphs. I. Bias, balance, and gains.
\newblock {\em J. Combin. Theory Ser. B}, 47:32 -- 52, 1989.




\end{thebibliography}


\end{document}




